Re: [PATCH 4/5] regulator: pwm: Add support for voltage linear equal steps

From: Laxman Dewangan
Date: Tue Mar 15 2016 - 02:58:16 EST

On Monday 14 March 2016 09:58 PM, Mark Brown wrote:
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On Sun, Mar 13, 2016 at 06:36:06PM +0530, Laxman Dewangan wrote:
On Saturday 12 March 2016 11:39 AM, Mark Brown wrote:
I can't see any reason why this would ever be preferable to just using
the flat linear range (you certainly haven't articulated one, you're
just stating it). This seems like you are bodging around a limited
consumer driver, you should fix the consumer to cope with regulators
with lots of voltages - PWM regulators aren't the only ones with high
resolution steps.
The requirement is to have perfect linear steps interms of the period/pulse
time of PWM without loosing any voltage.
Continuous mode is pretty much near to what you said but here we are loosing
the perfect step as this divides the periods to 100 parts and then set
Could you be more specific about what the issue is? We've hopefully got
errors of less than 1% in the values here...

If I use the continuous mode of PWM regulator then the calculation for PWM pulse ON time(duty_pulse)
done as:
duty_cycle = ((requested - minimum) * 100) / voltage_range.

duty_pulse = (pwm_period/100) * duty_cycle

This leads to the calculation error if we have the requested voltage where accurate pulse time is possible. For example: Let's have following case
voltage range is 800000uV to 1350000uV.
pwm-period = 1550ns (1ns time is 1mV).
Requested 900000uV.

duty_cycle = ((900000uV - 800000uV) * 100)/ 1550000
= 6.45 but we will get 6 due to integer division.

duty_pulse = (1550/100) * 6 = 90 pulse time.

90 pulse time is equivalent to 90mV and this gives us pulse time equivalent
to 890000uV instead of 900000uV.

If new mode is not accpetable then need to enhance the existing continuous
mode like before scaling for 100% of period, first look if we get the
perfect pulse time of of PWM period and if it is there then use this direct
instead of converting required voltage to 100% scale and then back
calculating duty time.
That seems a lot better,

You mean using the continuous mode only.

If I add following logic then also it resolve the issue:
if (((req_uV - min_uV) * pwm_period) % voltage_range == 0)
duty_pulse = ((req_uV - min_uV) * pwm_period) / voltage_range;
existing_continuous mode calculation.

So on above example:
duty_pulse = ((900000uV - 800000uV) * 1550)/1550000)
= 100

and this is equivalent to 100mV and so final voltage is
(800000 + 100000) = 900000uV which is same as requested,